Question: $f(x)=3{{x}^{3}}-{{x}^{2}}-4x+7$ What is the coefficient for the term containing $(x-1)^3$ in the Taylor polynomial, centered at $x=1$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $3$ (Choice C) C $6$ (Choice D) D $18$
Solution: Each term of a Taylor Series Polynomial centered at $~x=1~$ is in the form of $\frac{{{f}^{(n)}}(1){{(x-1)}^{n}}}{n!}\,$. We need the third derivative of $~f\left( x \right)~$ evaluated at $~x=1\,$. $f(x)=3{{x}^{3}}-{{x}^{2}}-4x+7$ $ f\,^\prime(x)=9{{x}^{2}}-2x-4$ $f\,^{\prime\prime}(x)=18x-2$ $f\,^{\prime\prime\prime}(x)=18$ Hence, $f\,^{\prime\prime\prime}\left( 1 \right)=18\,$. Therefore, the coefficient of the term containing $~{{\left( x-1 \right)}^{3}}~$ is $~\frac{18}{3!}=\frac{18}{6}=3\,$.